Answer: the pH of the solution is 4.52
Explanation:
Consider the weak acid as Ha, it is dissociated as expressed below
HA H⁺ + A⁻
the Henderson -Haselbach equation can be expressed as;
pH = pKa + log( [A⁻] / [HA])
the weak acid is dissociated into H⁺ and A⁻ ions in the solution.
now the conjugate base of the weak acid HA is
HA(aq) {weak acid} H⁺(aq) + A⁻(aq) {conjugate base}
so now we calculate the value of Kₐ as well as pH value by substituting the values of the concentrations into the equation;
pKₐ = -logKₐ
pKₐ = -log ( 7.4×10⁻⁵ )
pKₐ = 4.13
now thw pH is
pH = pKₐ + log( [A⁻] / [HA])
pH = 4.13 + log( [0.540] / [0.220])
pH = 4.13 + 0.3899
pH = 4.5199 = 4.52
Therefore the pH of the solution is 4.52
CAN YOU HELP ME ANWSER MINE ILL HELP YOU
Explanation:
Let us assume that total mass of the solution is 100 g. And, as it is given that acetic acid solution is 12% by mass which means that mass of acetic acid is 12 g and 88 g is the water.
Now, calculate the number of moles of acetic acid as its molar mass is 60 g/mol.
No. of moles =
= 
= 0.2 mol
Molarity of acetic acid is calculated as follows.
Density = 
1 g/ml = 
volume = 100 ml
Hence, molarity = 
= 
= 2 mol/l
As reaction equation for the given reaction is as follows.

So, moles of NaOH = moles of acetic acid
Let us suppose that moles of NaOH are "x".
(as 1 L = 1000 ml)
x = 20 L
Thus, we can conclude that volume of NaOH required is 20 ml.
Answer:
18.22874999999973
I recommend you to round the nearest 1 d.p
Explanation:
<em>h</em><em>a</em><em>v</em><em>e</em><em> </em><em>a</em><em> </em><em>g</em><em>r</em><em>e</em><em>a</em><em>t</em><em> </em><em>d</em><em>a</em><em>y</em><em>!</em>